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\author{学号 \underline{\hspace{4cm}} \hspace{1cm} 姓名 \underline{\hspace{4cm}} }
\title{复变函数练习 3.2 - 柯西积分定理 }
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\date{2024 年 4 月 15 日}
%\date{March 9, 2021}

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\begin{document}

\maketitle

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\begin{enumerate}

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\item  %Problem 01
%（柯西积分定理）
设函数 $f(z)$ 在复平面的单连通区域 $D$ 内解析（定义为 $f'(z)$ 处处存在），设 $f'(z)$ 在 $D$ 内连续，设 $C$ 为 $D$ 内任意一条分段光滑闭曲线。使用柯西-黎曼方程和格林定理，证明柯西积分定理 
$$\int_C f(z)dz=0. $$ 

\vspace{0.0cm}

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\item  %Problem 02
设函数 $f(z)$ 在复平面的单连通区域 $D$ 内解析，使用柯西积分定理，证明下述积分与$D$ 内连接起点 $z_0$ 和终点 $z_1$ 的曲线无关，$$\int_{z_1}^{z_2} f(z)dz. $$ 

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\item  %Problem 03
**使用古尔萨的方法，分别对积分路径是三角形、简单闭折线、和分段光滑闭曲线的情形，证明柯西积分公式。
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\item  %Problem 04
设函数 $f(z)$ 在单连通区域 $D$ 内解析，设起点 $z_0\in D$ 固定，设动点 $z\in D$, 证明积分
$$F(z) = \int_{z_0}^z f(\zeta)d\zeta$$
在区域 $D$ 内解析，而且 $F'(z)=f(z)$. 
  
\vspace{0.0cm}

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\item  %Problem 05
设 $D$ 是单连通区域 $-\pi< \arg z < \pi$, %函数 $\ln z$ 是 $f(z)=\frac{1}{z}$ 的一个原函数，
设 $z\in D$, 计算积分 $$\int_1^z \zeta^{-1}d\zeta. $$

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\item  %Problem 06
计算积分 
$$(1) \int_0^{\pi i} z\cos (z^2) dz, \hspace{0.5cm} (2) \int_0^{i} z\cos (z) dz $$

\vspace{0.0cm}

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\item  %Problem 07
沿着圆弧 $|z|=1$ 在区域 $\textrm{Im}(z)\ge 0, \textrm{Re}(z)\ge 0$ 的部分，计算积分
$$\int_1^i \frac{\ln (z+1)}{z+1}dz. $$

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\item  %Problem 08
设 $0<r<1$, 计算积分 $$\int_{|z|=r} \ln(1+z)dz. $$ 

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\item  %Problem 09
设 $C$ 是圆弧 $|z|=3$ 在区域 $\mathrm{Re}(z)\ge 0$ 的部分，设起点为 $-3i$, 终点为 $3i$, 计算积分 $$\int_C \frac{dz}{z^2}. $$

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\item  %Problem 10
取多值函数 $\sqrt{z}$ 的 $\sqrt{1}=-1$ 的单值分支，计算积分 $$\int_{|z-1|=1}\sqrt{z}dz. $$

\vspace{0.0cm}

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\item  %Problem 11
设 $C$ 为包含圆周 $|z|=1$ 的任意正向简单闭曲线，计算积分 $$\int_C \frac{2z-1}{z^2-z}dz. $$

\vspace{0.0cm}

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\item  %Problem 12
设区域 $G=\mathbb{C}-\{0\}$, 证明变上限积分
 $$\int_1^z \frac{d\zeta}{\zeta} = \mathrm{Ln}(z). $$ 

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\item  %Problem 12
设区域 $G=\mathbb{C}-\{0\}$, 计算变上限积分 
 $$\int_1^z \frac{d\zeta}{\zeta^2}.$$ 

\vspace{0.0cm}

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\item  %Problem 13
设 $C$ 是单位圆周 $|z|=1$, 计算积分 
$$(1) \int_C \frac{dz}{\cos(z)}, \hspace{0.5cm} 
(2) \int_C \frac{dz}{z^2+2z+2}, \hspace{0.5cm} 
(3) \int_C \frac{e^zdz}{z^2+5z+6}, \hspace{0.5cm} 
(4) \int_C z\cos(z^2)dz. 
$$

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\item  %Problem 14
计算积分 $$(1) \int_{-2}^{-2+i} (z+2)^2dz, \hspace{0.5cm} (2) \int_0^{\pi+2i} \cos(z/2)dz. $$

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\end{enumerate}


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\end{document}

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